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Modeling Quadratic Functions
Using Desmos
A farmer collected the data below, which shows crop yields (y1) for the various amounts of fertilizer (x1) used.
~ Plot these values using a Desmos Table
{(0,4), (0,6), (5, 10), (5,7), (10, 12), (10, 10), (15, 15), (15,17), (20, 18), (20, 21), (25, 20), (25, 21), (30, 21), (30, 22), (35, 21), (35, 20), (40, 19), (40, 20)}
~ You will need to change the domain and range for the graph
* Determine the domain by looking at the x1 values.
What is the smallest value, and what is the largest value?
* Determine the range by looking at the y1 values.
What is the smallest value, and what is the largest value?
~ Click on the wrench to change the domain and range.
Make sure you can see all of the plotted data points.
~ Find the quadratic function model by using the following:
y1 ~ a(x1 - h)2 + k
~ A graph of the best fit function will be graphed.
~ The values for a, h, and k will be generated by Desmos
a = -0.017121 h = 31.438 k = 20.816
~ Plug in the above values into the quadratic equation form:
y = a(x - h)2 + k
y = -0.017121 (x - 31.438)2 + 20.816
~ Determine the optimal amount of fertilizer to apply.
Use the following quadratic form y = a(x - h)2 + k
~ Vertex coordinates are (h, k)
These are the coordinates (x, y) of the vertex of the graph (above graph)
y = -0.017121(x - 32.059)2 + 20.932
Notice the value of "a" is negative, therefore, the graph's vertex is the maximum.
The highest (or optimal amount) is h = 32.059 pounds of fertilizer per 100 square feet.
~ Predict crop yield when the optimal amount of fertilizer is applied.
Using the above graph and quadratic model, the crop yield will be k = 20.932 bushels.
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